At what value of x does the function f(x) = x | vedclass

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(C) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\ln y = x \ln x$.Differentiating with respect to $x$,we have $\frac{1}{y} \frac{

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$x = e^{-1}$

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(C) Let $y = x^x$. Taking the natural logarithm on both sides,we get $\ln y = x \ln x$.Differentiating with respect to $x$,we have $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$.Thus,$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$.For critical points,set $\frac{dy}{dx} = 0$. Since $x^x > 0$ for $x > 0$,we have $\ln x + 1 = 0$,which implies $\ln x = -1$,so $x = e^{-1}$.To check for the minimum,we find the second derivative or use the first derivative test. For $x  e^{-1}$,$\ln x > -1$,so $\frac{dy}{dx} > 0$.Since the derivative changes sign from negative to positive at $x = e^{-1}$,the function attains its minimum value at $x = e^{-1}$.

At what value of $x$ does the function $f(x) = x^x$ $(x > 0)$ attain its minimum value?

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